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  • Calculating Nitrogen Dioxide (NO2) Required for Nitric Oxide (NO) Production
    Here's how to solve this problem:

    1. Balanced Chemical Equation:

    The balanced chemical equation for the reaction is:

    3 NO₂ (g) + H₂O (l) → 2 HNO₃ (aq) + NO (g)

    2. Mole-to-Mole Ratio:

    From the balanced equation, we see that 3 moles of NO₂ react to produce 1 mole of NO.

    3. Convert Molecules to Moles:

    * We have 5.00 x 10²² molecules of NO.

    * Avogadro's number tells us there are 6.022 x 10²³ molecules in one mole.

    * Moles of NO = (5.00 x 10²² molecules) / (6.022 x 10²³ molecules/mol) = 0.0830 mol NO

    4. Calculate Moles of NO₂:

    * Using the mole ratio from the balanced equation:

    Moles of NO₂ = 0.0830 mol NO * (3 mol NO₂ / 1 mol NO) = 0.249 mol NO₂

    5. Convert Moles of NO₂ to Grams:

    * The molar mass of NO₂ is 14.01 g/mol (N) + 2 * 16.00 g/mol (O) = 46.01 g/mol

    * Grams of NO₂ = 0.249 mol NO₂ * 46.01 g/mol = 11.4 g NO₂

    Therefore, you need approximately 11.4 grams of nitrogen dioxide to react with water to produce 5.00 x 10²² molecules of nitrogen monoxide.

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